Trilinear polarity

Construction of a trilinear polar of a point P

  Given triangle △ABC

  Cevian triangle △DEF of △ABC from P

  Cevian lines which intersect at P

  Constructed trilinear polar (line XYZ)

Let △ABC be a plane triangle and let P be any point in the plane of the triangle not lying on the sides of the triangle. Briefly, the trilinear polar of P is the axis of perspectivity of the cevian triangle of P and the triangle △ABC.

In detail, let the line AP, BP, CP meet the sidelines BC, CA, AB at D, E, F respectively. Triangle △DEF is the cevian triangle of P with reference to triangle △ABC. Let the pairs of line (BC, EF), (CA, FD), (DE, AB) intersect at X, Y, Z respectively. By Desargues' theorem, the points X, Y, Z are collinear. The line of collinearity is the axis of perspectivity of triangle △ABC and triangle △DEF. The line XYZ is the trilinear polar of the point P.[1]

The points X, Y, Z can also be obtained as the harmonic conjugates of D, E, F with respect to the pairs of points (B, C), (C, A), (A, B) respectively. Poncelet used this idea to define the concept of trilinear polars.[1]

If the line L is the trilinear polar of the point P with respect to the reference triangle △ABC then P is called the trilinear pole of the line L with respect to the reference triangle △ABC.

Let the trilinear coordinates of the point P be p : q : r. Then the trilinear equation of the trilinear polar of P is[3]

${\displaystyle {\frac {x}{p}}+{\frac {y}{q}}+{\frac {z}{r}}=0.}$

Construction of a trilinear pole of a line XYZ

  Given trilinear polar (line XYZ)

  Given triangle △ABC

  Cevian triangle △UVW of △ABC from XYZ

  Cevian lines, which intersect at the trilinear pole P

Let the line L meet the sides BC, CA, AB of triangle △ABC at X, Y, Z respectively. Let the pairs of lines (BY, CZ), (CZ, AX), (AX, BY) meet at U, V, W. Triangles △ABC and △UVW are in perspective and let P be the center of perspectivity. P is the trilinear pole of the line L.

Some of the trilinear polars are well known.[4]

• The trilinear polar of the centroid of triangle △ABC is the line at infinity.
• The trilinear polar of the symmedian point is the Lemoine axis of triangle △ABC.
• The trilinear polar of the orthocenter is the orthic axis.
• Trilinear polars are not defined for points coinciding with the vertices of triangle △ABC.

Animation illustrating the fact that the locus of the trilinear poles of a pencil of lines passing through a fixed point K is a circumconic of the reference triangle.

Let P with trilinear coordinates X : Y : Z be the pole of a line passing through a fixed point K with trilinear coordinates x₀ : y₀ : z₀. Equation of the line is

${\displaystyle {\frac {x}{X}}+{\frac {y}{Y}}+{\frac {z}{Z}}=0.}$

Since this passes through K,

${\displaystyle {\frac {x_{0}}{X}}+{\frac {y_{0}}{Y}}+{\frac {z_{0}}{Z}}=0.}$

Thus the locus of P is

${\displaystyle {\frac {x_{0}}{x}}+{\frac {y_{0}}{y}}+{\frac {z_{0}}{z}}=0.}$

This is a circumconic of the triangle of reference △ABC. Thus the locus of the poles of a pencil of lines passing through a fixed point K is a circumconic E of the triangle of reference.

It can be shown that K is the perspector[5] of E, namely, where △ABC and the polar triangle[6] with respect to E are perspective. The polar triangle is bounded by the tangents to E at the vertices of △ABC. For example, the Trilinear polar of a point on the circumcircle must pass through its perspector, the Symmedian point X(6).

• Geometrikon page : Trilinear polars
• Geometrikon page : Isotomic conjugate of a line